Proof. This is clear since $U(\mathfrak {g}) \cong U(\bar {\mathfrak {g}})$ as $\mathfrak {g}$ -modules with respect to the adjoint action. By taking $\mathfrak {g}$ -invariants, we get (3).

Proof. For any $\mathfrak {T}$ -module X, we have

$$ \begin{align*} \operatorname{\mathrm{Hom}}_{\mathfrak{T}}\left( Q(\lambda),X \right) &\cong \operatorname{\mathrm{Hom}}_{\mathfrak{T}}\left( \operatorname{\mathrm{Ind}}_{\mathfrak{g}}^{\mathfrak{T}} L(\lambda),X \right) \\ & \cong \operatorname{\mathrm{Hom}}_{\mathfrak{g}}\left( L(\lambda),X \right) \\ & \cong \operatorname{\mathrm{Hom}}_{\mathfrak{g}}\left( L(0) , \operatorname{\mathrm{Hom}}_{\mathbb{C}}\left( L(\lambda),X \right) \right) \\ & \cong \operatorname{\mathrm{Hom}}_{\mathfrak{T}}\left( \operatorname{\mathrm{Ind}}_{\mathfrak{g}}^{\mathfrak{T}} L(0), \operatorname{\mathrm{Hom}}_{\mathbb{C}}\left( L(\lambda),X \right) \right) \\ & \cong \operatorname{\mathrm{Hom}}_{\mathfrak{T}}\left( Q(0) \mathbin{\mathop{\otimes}\limits_{\mathbb{C}}} L(\lambda), X \right), \end{align*} $$

where we have used the fact that the induction is left adjoint to the forgetful functor, as well as the standard hom-tensor duality for Lie algebra modules. This finishes the proof.

Proof. The module $Q(0)$ is generated by $1 \otimes 1$ by construction, so any endomorphism of $Q(0)$ is uniquely determined by the image of $1 \otimes 1$ . Denote this image by $u \otimes 1$ , for some $u \in U(\bar {\mathfrak {g}})$ . The element $u\otimes 1$ generates the trivial $\mathfrak {g}$ -submodule (since $1\otimes 1$ does), so u must commute with $\mathfrak {g}$ . Of course, u commutes with $\bar {\mathfrak {g}}$ . Hence, $u \in Z(\mathfrak {T}) \cap U(\bar {\mathfrak {g}}) = \overline {Z(\mathfrak {g})}$ .

Conversely, any $u \in \overline {Z(\mathfrak {g})}$ , being central, defines an endomorphism of $Q(0)$ . This endomorphism maps $1 \otimes 1$ to $u \otimes 1$ . The claim follows.

Proof.

1. (a) Suppose first that $\lambda =0$ . Then, as a $\mathfrak {g}$ -module, $Q(0)$ is isomorphic to $U(\mathfrak {g})$ with respect to the adjoint action. Taking the $\chi $ -component of $Q(0)$ corresponds to factoring $U(\mathfrak {g})$ by the ideal generated by the corresponding central character of $Z(\mathfrak {g})$ . From Kostant’s theorem (see [Reference Jantzen32, Section 3.1]), it follows that $Q(0)$ decomposes as direct sum of finite-dimensional $\mathfrak {g}$ -submodules, and that $[Q(0,\chi ) \colon L(\mu )] = \dim L(\mu )_0$ . The general statement now follows from (5).

   Note that the value in (6) is finite, since, for fixed $\mu $ and $\lambda $ , the value $l^{\mu }_{\nu ,\lambda }$ is nonzero only for finitely many $\nu $ .

2. (b) This follows from Schur’s Lemma by adjunction.

Proof. By Proposition 8, the module $Q(0,\chi )$ has a unique occurrence of $L(0)$ , and is generated by it. Therefore, the sum all its submodules not containing $L(0)$ as a composition factor is the unique maximal submodule; denote it by N. It follows that is the unique simple quotient of $Q(0,\chi )$ .

Suppose now that V is finite-dimensional. The nilradical $\bar {\mathfrak {g}}$ must act trivially on it. Because of simplicity, we must have $V=L(0)$ , which is a contradiction.

Proof. The first claim is clear. The second one follows from the PBW basis in $U(\mathfrak {g})$ and the relation $\bar {h}^2 = -4\bar {f} \bar {e} + \chi $ in the quotient.

The decomposition in the third claim is given by Kostant’s theorem (see [Reference Jantzen32, Section 3.1]). Since $\bar {e}^k$ is of weight $2k$ and annihilated by e, it must be a highest weight vector of a $\mathfrak {g}$ -submodule isomorphic to $L(2k)$ , which, we know, occurs uniquely in $Q(0,\chi )$ .

The last claim follows from the definitions by a direct calculation.

Proof. Assume $\chi \neq 0$ , and let $V \subseteq Q(0,\chi )$ be nonzero submodule. Take k to be the smallest nonnegative integer such that $L(2k) \subseteq V$ . If $k=0$ , then $V = Q(0,\chi )$ , since $L(0)$ generates $Q(0,\chi )$ , and we are done. So, let us assume now $k \geq 1$ . We have $\bar {e}^k \in V$ , so if we find an element from $U(\mathfrak {T})$ that maps $\bar {e}^k \in V$ to $\bar {e}^{k-1}$ , we will get a contradiction. That element can be taken as $\frac {1}{k\chi }(4k \bar {f} - \bar {h}f)$ , namely:

$$ \begin{align*} (4k \bar{f} - \bar{h}f) \circ \bar{e}^k &= 4k\bar{f}\bar{e}^k - \bar{h}[f,\bar{e}^k] \\ &= 4k\bar{f}\bar{e}^k + k \bar{h}^2\bar{e}^{k-1} \\ &= 4k\bar{f}\bar{e}^k + k (-4\bar{f} \bar{e} + \chi)\bar{e}^{k-1} \\ &= k\chi \bar{e}^{k-1}. \end{align*} $$

We conclude that $Q(0,\chi )$ is simple.

For the converse, assume $\chi = 0$ . We will show that for any $k \geq 0$ , the subspace $Q_k := \oplus _{t \geq k} L(2t)$ is a submodule. From this, the theorem will follow.

Let us first prove that $L(2k)$ is equal to the span of $\big \{\bar {f}^i \bar {h}^\epsilon \bar {e}^j \colon \epsilon \in \{0,1\}, i+\epsilon + j = k \big \}$ . This set contains $\bar {e}^k$ , so it is enough to see that it is stable under f. We calculate the two cases whether $\epsilon $ is $0$ or $1$ separately:

$$ \begin{align*} f \circ \bar{f}^i \bar{e}^j &= \bar{f}^i [f,\bar{e}^j] = - j \bar{f}^i \bar{h} \bar{e}^{j-1}, \\[.5em] f \circ \bar{f}^i \bar{h} \bar{e}^j &= \bar{f}^i [f,\bar{h}]\bar{e}^j + \bar{f}^i \bar{h} [f,\bar{e}^j] \\ &= 2\bar{f}^{i+1} \bar{e}^j -j \bar{f}^i \bar{h}^2 \bar{e}^{j-1} \\ &= 2\bar{f}^{i+1} \bar{e}^j +4j \bar{f}^{i+1} \bar{e}^{j} \\ &= (4j+2)\bar{f}^{i+1} \bar{e}^j. \end{align*} $$

From this description of $L(2k)$ , one easily checks that $\bar {f}, \bar {h},\bar {e}$ map $L(2k)$ to $L(2k+2)$ . From this, it follows that $Q_k$ is a submodule.

Proof. From the explicit description of generators of the center, we get a system of equations

$$ \begin{align*} \begin{cases} \lambda_2(\lambda_1+2)=\lambda_2'(\lambda_1'+2) \\ \lambda_2^2 = (\lambda_2')^2 \end{cases}\!\!, \end{align*} $$

which is easily solved.

Proof. Denote by $v_\lambda $ a basis element of $\mathbb {C}_\lambda $ . Then, $\Delta (\lambda )$ has a basis of weight vectors $\{ f^i \bar {f}^j v_\lambda \colon i,j \geq 0\}$ . A direct computation (with help of [Reference Humphreys30, Lemma 21.2] and its Takiff analogue, alternatively use [Reference Cai, Shen and Zhang12, Lemma 2.1]) shows that

(8) $$ \begin{align} e \cdot f^i \bar{f}^j v_\lambda &= [e , f^i] \bar{f}^j v_\lambda + f^i [e,\bar{f}^j ] v_\lambda \\ \nonumber &= i(\lambda_1 - i -2j +1) f^{i-1} \bar{f}^j v_\lambda + j \lambda_2 f^i \bar{f}^{j-1} v_\lambda. \end{align} $$

This implies that the required filtration is given by the degree of $\bar {f}$ . The subquotients are given by the span of $\{ f^i \bar {f}^k v_\lambda \colon i \geq 0\}$ , which is clearly isomorphic to $\Delta ^{\mathfrak {g}}(\lambda _1 - 2k)$ .

If $\lambda _2 =0$ , it is clear that the span of $\{ f^i \bar {f}^k v_\lambda \colon i \geq 0\}$ , k fixed, is a $\mathfrak {g}$ -submodule. If $\lambda _1 \not \in \mathbb {Z}_{\geq 0}$ , then there are no possible nontrivial extensions between $\Delta ^{\mathfrak {g}}(\lambda _1 - 2k)$ , $k \geq 0$ ; hence $\Delta (\lambda )$ splits as a direct sum of these.

Suppose now $\lambda _2 \neq 0$ and $\lambda _1 \in \mathbb {Z}_{\geq 0}$ . Fix $\mu \in \{0,1,\ldots ,\lambda _1\}$ of the same parity as $\lambda _1$ . It is enough to show that $\Delta ^{\mathfrak {g}}(-\mu - 2)$ is not a $\mathfrak {g}$ -submodule of $\Delta (\lambda )$ . Suppose it is. Its highest weight vector $v_{-\mu -2}$ must be a nontrivial linear combination of $f^i \bar {f}^j v_\lambda $ with $i+ j = \frac {\lambda _1+\mu }{2}+1 =:t$ , with a nonzero coefficient by $\bar {f}^{t} v_\lambda $ .

From (8), it follows that the matrix of e in bases $f^{i} \bar {f}^{t-i} v_\lambda $ , $i=0,\ldots ,t$ , and $f^{i} \bar {f}^{t-1-i} v_\lambda $ , $i=0,\ldots ,t-1$ , has the form

$$ \begin{align*} \begin{pmatrix} \ast & \ast & 0 & \ldots & 0 & 0 & 0 \\ 0 & \ast & \ast & \ldots & 0 & 0 & 0 \\ 0 & 0 & \ast & \ddots & \vdots & \vdots & \vdots \\ \vdots & \vdots & \vdots & \ddots & \ast & 0 & 0 \\ 0 & 0 & 0 & \ldots & \ast & \ast & 0 \\ 0 & 0 & 0 & \ldots & 0 & \ast & \ast \end{pmatrix}, \end{align*} $$

with all $\ast $ nonzero, except the one on the position $(\mu +1,\mu +2)$ , where we have zero (because the bracket in (8) is zero for $f^{\mu +1} \bar {f}^{\left (\frac {\lambda _1-\mu }{2} \right )} v_\lambda $ ). From this, it follows that e cannot annihilate $v_{-\mu -2}$ , a contradiction.

Proof. In the same way as for the semisimple case (see, e.g., [Reference Humphreys31, Section 6.3]), one sees that the left-hand side has a filtration with subquotients equal to the summands on the right-hand side. But these subquotients have different central characters, which follows from Lemma 15, so they split.

Proof. The set in the first claim is a generating set for $U_{(x)}$ , which follows from PBW and the properties of Ore localization. But this set is also linearly independent, since for its any finite subset, the multiplication from the left by $x^m$ for some large m produces a linearly independent set in $U \leq U_{(x)}$ . This proves the first claim, and the second claim follows from it.

Proof. There is an isomorphism $U_{(x)} \otimes _{U} (M \otimes L) \to (U_{(x)} \otimes _{U} M) \otimes L$ given by

$$ \begin{align*} x^{-n} \otimes (m \otimes v) \mapsto \sum_{k\geq 0} (-1)^k {n+k-1 \choose k} (x^{-n-k} \otimes m) \otimes x^k v, \end{align*} $$

with the inverse given by $(x^{-n} \otimes m) \otimes v \mapsto x^{-ar} \otimes \sum _{k \geq 0} {ar \choose k}( x^{ar-n-k} m \otimes x^kv)$ , where $r,a \in \mathbb {Z}_{>0}$ are chosen, so that $x^r$ annihilates L and $(r-1)a \geq n$ . This is proved in [Reference Deodhar15, Theorem 3.1] and [Reference Andersen and Stroppel2, Theorem 3.2] for the semisimple case, but the proof is analogous in general. In proving that these maps compose to the identity, the following combinatorial formula is helpful: $\sum _{k=0}^n (-1)^k{a \choose n-k} {b+k \choose k} = {a-b-1 \choose n}$ .

One can check that these isomorphisms preserve the canonical images of $M \otimes L$ in both sides (see (9)), so they induce the required isomorphisms on the quotients.

Proof. Since x is assumed to be $\operatorname {\mathrm {ad}}$ -nilpotent, the claim follows from the formula in [Reference Humphreys30, Lemma 21.4].

Proof. Because of Lemma 19, it is enough to show that ${}^{e}(M \otimes L) = ({}^{e}\!M) \otimes L$ for $\mathfrak {g}$ -modules M and L with $L=L(\mu )$ simple finite-dimensional. This is proved in [Reference Deodhar15, Corollary 3.2], but we also give a proof for the sake of completeness.

The inclusion $({}^{e}\!M) \otimes L \subseteq {}^{e}(M \otimes L)$ is trivial. For the converse, denote by v the lowest weight vector of L. Then, $v, e v, \ldots , e^{\mu } v$ is a basis for L. Take a general element $m = \sum _{i=0}^\mu m_i \otimes e^i v \in {}^{e}(M \otimes L)$ , and observe that for $n> \mu $ , we have

$$ \begin{align*} e^n \cdot m &= \sum_{i=0}^\mu \sum_{j=0}^{\mu-i} {n \choose j}e^{n-j} m_i \otimes e^{i+j} v \\ &= \sum_{i=0}^\mu \left( e^n m_i + \sum_{j=0}^{i-1} {n \choose i-j}e^{n+j-i} m_j \right) \otimes e^{i} v. \end{align*} $$

For a fixed i, the vectors inside the big brackets must span a finite-dimensional space when n varies. From this, and an induction on i, it follows that $e^n m_i$ span a finite-dimensional space, hence $m \in ({}^{e}\!M) \otimes L$ .

Proof. Lemma 16, the fact that the functor $\operatorname {\mathrm {\mathbf {EA}}}$ commutes with the forgetful functor from $\mathfrak {T}$ -modules to $\mathfrak {g}$ -modules, and Example 23 imply (12).

From Lemma 18, we have a basis for $S_f \mathbin {\mathop {\otimes }\limits _{U}} \Delta (\lambda )$ consisting of $f^{-i} \bar {f}^j v_\lambda $ , for $i \geq 1$ and $j \geq 0$ . Since the lowest weight vector of a $\mathfrak {g}$ -type $L(\mu )$ ( $\mu $ of the same parity as $\lambda _1$ ) inside $\operatorname {\mathrm {\mathbf {EA}}}(\Delta (\lambda ))$ must be annihilated by f, it must be (up to scalar) equal to $f^{-1} \bar {f}^t v_\lambda $ , where $t = \frac {\mu +\lambda _1}{2}+1$ .

Now, we will prove that $\operatorname {\mathrm {\mathbf {EA}}}(\Delta (\lambda ))$ is simple. Let V be its nonzero submodule, and suppose it contains $L(\mu )$ for some $\mu $ from (12). By applying $\bar {f}$ on $f^{-1} \bar {f}^t v_\lambda $ , we get that $L(\mu +2k) \subseteq V$ , for all $k \geq 0$ .

To prove that $V=\operatorname {\mathrm {\mathbf {EA}}}(\Delta (\lambda ))$ , it is enough to assume $\mu> |\lambda _1+2|$ and to find an element in $U(\mathfrak {T})$ that maps $f^{-1} \bar {f}^t v_\lambda $ to $f^{-1} \bar {f}^{t-1} v_\lambda $ .

In addition to (11), we will use the following commutation relations, whose proofs are analogous to the ones for (11):

(13) $$ \begin{align} &[\bar{h},f^{-k}] = 2k f^{-k-1} \bar{f}, \\ \nonumber &[\bar{e},f^{-k}] = - kf^{-k-1} \bar{h} - k(k+1)f^{-k-2} \bar{f}. \end{align} $$

From this, we have:

$$ \begin{align*} e \cdot f^{-1} \bar{f}^t v_\lambda &= [e,f^{-1}] \bar{f}^t v_\lambda + f^{-1} [e,\bar{f}^t] v_\lambda \\ &= - f^{-2} (h+2) \bar{f}^t v_\lambda + t \lambda_2 f^{-1} \bar{f}^{t-1} v_\lambda \\ &= \mu f^{-2} \bar{f}^t v_\lambda + t \lambda_2 f^{-1} \bar{f}^{t-1} v_\lambda, \\[.5em] \bar{h}e \cdot f^{-1} \bar{f}^t v_\lambda &= \mu \bar{h} f^{-2} \bar{f}^t v_\lambda + t \lambda_2 \bar{h} f^{-1} \bar{f}^{t-1} v_\lambda \\ &= \mu [\bar{h}, f^{-2}] \bar{f}^t v_\lambda + \mu \lambda_2 f^{-2} \bar{f}^t v_\lambda + t\lambda_2 [\bar{h},f^{-1}] \bar{f}^{t-1} v_\lambda + t\lambda_2^2 f^{-1} \bar{f}^{t-1} v_\lambda \\ &= 4\mu f^{-3} \bar{f}^{t+1} v_\lambda + (\lambda_1 + 2\mu + 2)\lambda_2 f^{-2} \bar{f}^{t} v_\lambda + t\lambda_2^2 f^{-1} \bar{f}^{t-1} v_\lambda, \\[.5em] 2\mu \bar{e} \cdot f^{-1} \bar{f}^t v_\lambda &= 2\mu[\bar{e}, f^{-1}] \bar{f}^t v_\lambda \\ &= -4\mu f^{-3} \bar{f}^{t+1} v_\lambda - 2\mu\lambda_2 f^{-2} \bar{f}^t v_\lambda. \end{align*} $$

From this, it follows that

$$ \begin{align*} (\bar{h}e - 2\mu \bar{e}) \cdot f^{-1} \bar{f}^t v_\lambda = (\lambda_1 + 2)\lambda_2 f^{-2} \bar{f}^{t} v_\lambda + t\lambda_2^2 f^{-1} \bar{f}^{t-1} v_\lambda. \end{align*} $$

Now, we claim that a nontrivial linear combination of e and $(\bar {h}e - 2\mu \bar {e})$ will map $f^{-1} \bar {f}^t v_\lambda $ to $f^{-1} \bar {f}^{t-1} v_\lambda $ . This is true, because the determinant

$$ \begin{align*} \begin{vmatrix} \mu & (\lambda_1+2)\lambda_2 \\ t\lambda_2 & t\lambda_2^2 \end{vmatrix} = \lambda_2^2(\mu+\lambda_1+2)(\mu-\lambda_1-2) \neq 0. \end{align*} $$

This finishes the proof of simplicity.

Proof. The first statement follows from Theorem 24.

It is clear that the functor $\operatorname {\mathrm {\mathbf {EA}}}$ preserves central character. So, the generators of the center C and $\bar {C}$ act as the scalars $n\lambda _2$ and $\lambda _2^2$ , respectively. From this, the second statement follows.

Proof. The first claim follows from Proposition 29, Corollary 28, and (5). The second claim follows from the first one by comparing both choices $\pm \lambda _2$ and central characters of the summands.

Proof. By Proposition 8(b), V is a quotient of $Q(n,\chi )$ .

If $\chi \neq 0$ , from Proposition 30 and Corollary 27, we see that the only possible choices with the correct minimal $\mathfrak {g}$ -type are $V(n,\lambda _2)$ or $V(n,-\lambda _2)$ .

If $\chi = 0$ , by the second part of Theorem 12, we see that the only possible simple quotients of $Q(n,0) \cong Q(0,0) \otimes L(n)$ are just finite-dimensional simple $\mathfrak {g}$ -modules with the trivial $\bar {\mathfrak {g}}$ -action.

Proof. Assume first that $V= Q(0,\chi )$ for $\chi \neq 0$ , and suppose we have a nonsplit short exact sequence $0 \to V \stackrel {i}{\hookrightarrow } M \stackrel {p}{\twoheadrightarrow } V \to 0$ . Denote by $1 \in V$ the generator from $L(0)$ and set $w=i(1) \in M$ . We can find $v \in M$ such that $p(v)=1$ and, in addition, such that v generates a trivial $\mathfrak {g}$ -submodule in M (note that the sequence must split in the category of $\mathfrak {g}$ -modules). By the universal property, there is a $\mathfrak {T}$ -homomorphism $f \colon Q(0) \to M$ , and the triangle below commutes:

The map f must be surjective, since otherwise its image would define a splitting of the short exact sequence. So, there is an element in $Q(0)$ that maps to w via f; by Lemma 11 and Proposition 7, such an element is necessarily of the form $p(\bar {C})$ for some polynomial p. Since the triangle above commutes, we must have $p(\chi )=0$ . Since $[M \colon L(0)] =2$ , we can take $p(\bar {C})=\bar {C}-\chi $ . From this, one can see that $\operatorname {\mathrm {Ker}} f$ is generated by $(\bar {C} - \chi )^2$ , that is, . This uniquely determines M. Conversely, one sees directly that such M defines a nonsplit self-extension of $Q(0,\chi )$ .

The general statement is obtained from this by translation functors, that is, tensoring extensions of $Q(0,\chi )$ by $L(n)$ and then taking the component with the correct central character (see Proposition 29). This functor defines a homomorphism of abelian groups

$$ \begin{align*}\operatorname{\mathrm{Ext}}^1(V(0,\lambda_2),V(0,\lambda_2)) \to \operatorname{\mathrm{Ext}}^1(V(n,\lambda_2),V(n,\lambda_2)). \end{align*} $$

In the same way, we get a homomorphism in the other direction. The fact that these homomorphisms compose to the identities on the $\operatorname {\mathrm {Ext}}^1$ groups is an easy application of the $5$ -lemma.

Proof. The only nonobvious thing to prove is if $u \in \operatorname {\mathrm {Ann}}(M)$ , then $u x^{-n} \otimes m =0$ , for all $n\geq 1$ and $m\in M$ .

By assumption, for any $u \in U$ , there exists $k_1>0$ such that

$$ \begin{align*} 0 = \left(\operatorname{\mathrm{ad}}(x)\right)^{k_1}(u) = \sum_{i=0}^{k_1} (-1)^{i}{k_1 \choose i} x^i u x^{k_1-i}, \end{align*} $$

so $x^{k_1} u = u_1 x$ for $u_1 := \sum _{i=0}^{k_1-1} (-1)^{k_1+i}{k_1 \choose i} x^i u x^{k_1-i-1}$ . If $u \in \operatorname {\mathrm {Ann}}(M)$ , then so is $u_1$ , since $\operatorname {\mathrm {Ann}}(M)$ is a two-sided ideal. We can inductively apply the same procedure on $u_1$ to get $k_2$ such that $x^{k_2} u_1 = u_2 x$ , and so on. Repeating this n times, we get $x^{k_{n}} u_{n-1} = u_n$ for some $u_n \in \operatorname {\mathrm {Ann}}(M)$ . From the construction, it follows that

$$ \begin{align*} x^{k_n+\cdots+k_1} \cdot ux^{-n} \otimes m = u_n \otimes m = 1 \otimes u_n m =0. \end{align*} $$

Since x acts injectively on M, the same is true for $U_{(x)} \mathbin {\mathop {\otimes }\limits _{U}} M$ . From this, it follows that $ux^{-n} \otimes m =0$ .

Proof. The first claim follows from Lemma 11(b) for $V(0,\lambda _2) \cong Q(0,\lambda _2^2)$ , and from Proposition 30 for general $V(n,\lambda _2)$ .

We prove the second claim also for $n=0$ , and again translate the result to the other cases. From Lemma 11(b), we get a basis for $W:=S_{\bar {e}} \mathbin {\mathop {\otimes }\limits _{U}} Q(0,\lambda _2^2)$ consisting of

$$ \begin{align*} (\bar{e})^{-k} \bar{f}^l \bar{h}^\epsilon, \text{ for } k>0, \ l \geq 0, \ \epsilon \in \{0,1\}, \end{align*} $$

where $\mathfrak {g}$ acts by the adjoint action, and $\bar {\mathfrak {g}}$ by the (commutative) multiplication. We denote this action of $\mathfrak {T}$ by $\circ $ .

Consider the following two elements in W:

$$ \begin{align*} w_\pm := (\bar{e})^{-1} \pm \frac{1}{\lambda_2}(\bar{e})^{-1}\bar{h}.\end{align*} $$

It is an easy calculation to see that $e \circ w_{\pm } = \bar {e} \circ w_{\pm } = 0$ , $h \circ w_{\pm } = -2 w_\pm $ , and $\bar {h} \circ w_\pm = \pm \lambda _2 w_{\pm }$ , from which it follows by the universal property of Verma modules that each $w_\pm $ generates a copy of $\Delta (-2,\pm \lambda _2)$ in W. Because of their simplicity, these submodules can only intersect trivially. By comparing the dimensions of weight spaces, we conclude that W cannot have any other composition factor, that is,

(14) $$ \begin{align} W \cong \Delta(-2,\lambda_2) \oplus \Delta(-2,-\lambda_2). \end{align} $$

In general, we calculate $S_{\bar {e}} \mathbin {\mathop {\otimes }\limits _{U}} Q(n,\lambda _2^2)$ in two ways and compare the results:

(15) $$ \begin{align} \nonumber S_{\bar{e}} {}\mathbin{\mathop{\otimes}\limits_{U}}{} & Q(n,\lambda_2^2) \cong S_{\bar{e}} \mathbin{\mathop{\otimes}\limits_{U}} \left( Q(0,\lambda_2^2) \mathbin{\mathop{\otimes}\limits_{}} L(n) \right) && \text{by (5)}, \\ \nonumber\cong & \left( S_{\bar{e}} \mathbin{\mathop{\otimes}\limits_{U}} Q(0,\lambda_2^2) \right) \mathbin{\mathop{\otimes}\limits_{}} L(n) && \text{by Lemma 19}, \\ \nonumber\cong & \, \big( \Delta(-2,\lambda_2) \oplus \Delta(-2,-\lambda_2) \big) \mathbin{\mathop{\otimes}\limits_{}} L(n) && \text{by (14)}, \\ \cong & \bigoplus_{k=0}^n \Delta(n-2-2k,\lambda_2) \oplus \bigoplus_{k=0}^n \Delta(n-2-2k,-\lambda_2) && \text{by Lemma 17}. \end{align} $$

On the other hand, by Proposition 30, we have

(16) $$ \begin{align} S_{\bar{e}} \mathbin{\mathop{\otimes}\limits_{U}} Q(n,\lambda_2^2) \cong \bigoplus_{k=0}^n S_{\bar{e}} \mathbin{\mathop{\otimes}\limits_{U}} V(n-2k,\lambda_2). \end{align} $$

By comparing the central characters (which are preserved under the localization) of the direct summands in (15) and (16), the claim (b) follows.

Proof. We need to prove that any product $x = x_1 x_2 \ldots x_k$ , where each $x_i$ belongs to the standard basis of $\mathfrak {T}$ , with the property that the number of i’s for which $x_i \in \{f,\bar {f}\}$ is equal to the number of j’s for which $x_j \in \{e,\bar {e}\}$ , can be generated by elements in S. We prove this by induction on k. If $x_1 x_2 \ldots x_k$ consists only of h and $\bar {h}$ , we are done. If not, chose some $x_i \in \{f,\bar {f}\}$ and $x_j \in \{e,\bar {e}\}$ , and assume without loss of generality $i<j$ . We commute them to the rightmost place:

$$ \begin{align*} x = (\underbrace{x_1 \ldots \hat{x_i} \ldots \hat{x_j} \ldots x_k}_{x'})(\underbrace{x_i x_j}_{\in S}) + \sum_t y_t,\end{align*} $$

where the factors with hat are omitted. It is clear from the commutation relations that $x'$ and all $y_t$ are products of the basis elements with the same property, but shorter. We are done by induction.

Proof. In the quotient above, by using (7), we can express $\bar {f}\bar {e}$ as a linear combination of $\bar {h}^2$ and $1$ , and also $\bar {f}e$ as a linear combination of $\bar {h}h$ , $\bar {h}$ , $f\bar {e}$ , and $1$ . Using this with Lemma 36, we see that the generators in the quotient are just h, $\bar {h}$ , $fe$ , and $f\bar {e}$ .

First, let us assume that $u=x_1 x_2 \ldots x_r$ , a product of these four generators in any order. Since h commutes with everything here, we can ignore it. Denote by $a=fe$ , $b=f\bar {e}$ , and $c=\bar {h}$ , $\chi _1 = \chi (C)$ and $\chi _2=\chi (\bar {C})$ . One can check that we have the following relations in the quotient:

(17) $$ \begin{align} [b,a] &= ac -hb, \end{align} $$

(18) $$ \begin{align} [c,a] &= 4b + hc + 2c - \chi_1, \end{align} $$

(19) $$ \begin{align} [c,b] &= \frac{1}{2}c^2 -\frac{1}{2}\chi_2, \end{align} $$

(20) $$ \begin{align} \bar{f}\bar{e} \cdot a &= -b^2 - hc^2 - \frac{3}{2}c^2 - hbc - 2bc + \frac{\chi_1}{2}b + \frac{\chi_1}{2}c + \frac{\chi_2}{2}. \end{align} $$

Suppose that $x_1 \neq a$ , but some $x_i =a$ , and assume i is minimal. Using the relations (17) and (18), we commute $x_{i-1}x_i = x_i x_{i-1} + [x_{i-1},x_i]$ . In this way, u becomes a sum of several monomials, each of which has either one a less, or have their most-left a one place closer to the most-left position. It follows that we can move this a to the most left part in a finite number of steps, that is, we can express u as a finite sum $u =\sum a y_t + \sum w_t$ , where $y_t$ is a finite product of a’s, b’s, and c’s, but has at least one a less than the original expression of u had, and $w_t$ is a finite product of b’s and c’s.

From (20) and the relation $4\bar {f}\bar {e} = \chi _2 - c^2$ , it follows that $\bar {f}\bar {e} \cdot u$ is a finite sum $\sum z_t$ , where each $z_t$ is a product of a’s, b’s, and c’s, but has at least one a less than the original expression of u had. By induction, for some k, we get that $(\bar {f}\bar {e})^k \cdot u$ is a finite sum of products of b’s and c’s.

Now, observe that any product of b’s and c’s can be expressed as a linear combination of standard monomials $b^i c^j$ , using the relation (19) and a very similar reasoning as before. The point is that a does not appear in $[c,b]$ in (19), so we will not end up in an infinite loop.

Finally, note that the argument is essentially the same if we started from u equal to a linear combination of products of the generators, instead of just one monomial.

Proof. We prove this for the annihilator of the Verma module $\Delta := \Delta (n-2,\lambda _2)$ . This is known from [Reference Bavula and Lu6, Proposition 6.1], but we present here a different and a more direct proof.

The inclusion $U \cdot \operatorname {\mathrm {Ker}} \chi \subseteq \operatorname {\mathrm {Ann}}(\Delta )$ is trivial. For the converse, recall that $\operatorname {\mathrm {Ann}}(\Delta )$ is stable under the adjoint action, so it is generated by its $U_0$ part. So, it is enough to prove

$$ \begin{align*} U_0 \cap \operatorname{\mathrm{Ann}}(\Delta) \subseteq U_0 \cdot \operatorname{\mathrm{Ker}} \chi. \end{align*} $$

To prove this, for any nonzero element , we want to find an element from $\Delta $ which is not annihilated by u. Because of Lemma 37, we can assume without loss of generality that

$$ \begin{align*} u = \sum_{k,l,m \geq 0} \alpha_{klm} h^k \left(f \bar{e}\right)^l \left(\bar{h}\right)^m, \end{align*} $$

with $\alpha _{klm} \in \mathbb {C}$ and only finitely many nonzero. Define a polynomial (with commutative variables) by the same scalars: $p(x,y,z)= \sum _{k,l,m \geq 0} \alpha _{klm} x^k y^l z^m \in \mathbb {C}[x,y,z]$ .

Denote by v the highest weight vector in $\Delta $ , by $\Delta _{q}$ the weight space in $\Delta $ of weight $n-2-2q$ , $q \geq 0$ , and recall that it has basis $f^i \bar {f}^{q-i} v$ , for $i=0,1,\ldots ,q$ . Similarly to (8), one can prove the following formulas for the action on $\Delta $ :

(21) $$ \begin{align} \nonumber h \cdot f^i \bar{f}^{q-i} v &= (n-2-2q) f^i \bar{f}^{q-i} v, \\ \nonumber f\bar{e} \cdot f^i \bar{f}^{q-i} v &= i \lambda_2 f^i \bar{f}^{q-i} v -i(i-1) f^{i-1} \bar{f}^{q-i+1} v, \\ \bar{h} \cdot f^i \bar{f}^{q-i} v &= \lambda_2 f^i \bar{f}^{q-i} v -2i f^{i-1} \bar{f}^{q-i+1} v. \end{align} $$

It follows that in this basis of $\Delta _q$ , the operator representing u is upper triangular, with the diagonal entries $p(n-2-2q,i \lambda _2,\lambda _2)$ , $i=0,\ldots ,q$ . We would like to find a basis element $f^i \bar {f}^{q-i} v$ , for which $p(n-2-2q,i \lambda _2,\lambda _2) \neq 0$ . However, a problem arises if $p(x,y,z)$ is divisible by $(z-\lambda _2)$ .

We claim that we can decompose

(22) $$ \begin{align} p(x,y,z) = \tilde{p}(x,y,z) \cdot (z-\lambda_2)^r, \end{align} $$

for some $r \geq 0$ , such that $\tilde {p}(n-2-2q,i \lambda _2,\lambda _2)$ is not identically zero for $(q,i) \in D$ , where $D \subseteq \mathbb {C}^2$ is any Zariski dense subset.

To prove this claim, write $p(x,y,z)= \sum _{j=0}^m p_j(x,y)(z-\lambda _2)^j$ . Suppose that this is zero when evaluated on $D \times \{\lambda _2\}$ for a Zariski dense set $D \subseteq \mathbb {C}^2$ . It follows that $p_0(x,y)=0$ (on $\mathbb {C}^2$ ), so $p(x,y,z)= p^{(1)}(x,y,z)(z-\lambda _2)$ , for a polynomial $p^{(1)}(x,y,z)$ of a strictly smaller total degree. If necessary, we continue to apply the same argument inductively on $p^{(1)}(x,y,z)$ , and so on, until we reach (22) with $\tilde {p}(x,y,\lambda _2)$ nonzero on some point in $(x,y) \in D$ . The number r is independent of D, since the set $\{(x,y) \colon \tilde {p}(x,y,\lambda _2) \neq 0\}$ is nonempty and Zariski open, hence intersects any Zariski dense set in $\mathbb {C}^2$ .

The claim is now proved, because the map $(q,i) \mapsto (n-2-2q,i \lambda _2)$ is an algebraic isomorphism $\mathbb {C}^2 \to \mathbb {C}^2$ . Here, it is crucial that $\lambda _2 \neq 0$ .

Write $\tilde {p}(x,y,z)= \sum _{k,l,m \geq 0} \tilde {\alpha }_{klm} x^k y^l z^m$ , and define $\tilde {u}= \sum _{k,l,m \geq 0} \tilde {\alpha }_{klm} h^k \left (f \bar {e}\right )^l \left (\bar {h}\right )^m$ . Then, it is also true that

$$ \begin{align*} u = \tilde{u} \cdot (\bar{h}-\lambda_2)^r, \end{align*} $$

since the monomials in u and $\tilde {u}$ have $\bar {h}$ on the most-right position, so no commuting of the variables is necessary.

There exists a pair $(q,i)$ from the cone $\{(q,i) \in \mathbb {Z}\times \mathbb {Z} \colon q\geq r, \ 0 \leq i \leq q-r \}$ (which is Zariski dense in $\mathbb {C}^2$ ), such that $\tilde {p}(n-2-2q,i \lambda _2,\lambda _2) \neq 0$ . Put $w:=f^{i+r} \bar {f}^{q-i-r}v \in \Delta _q$ . It follows from (21) that $(\bar {h}-\lambda _2)^r \cdot w = c \cdot f^{i} \bar {f}^{q-i}$ , for some constant $c \neq 0$ . From this, we have that

$$ \begin{align*} u \cdot w &= c \cdot \tilde{u} \cdot f^{i} \bar{f}^{q-i}v \\ &= c \cdot \tilde{p}(n-2-2q,i \lambda_2,\lambda_2) \cdot f^{i} \bar{f}^{q-i}v + \sum_{j=0}^{i-1} c_j f^{j} \bar{f}^{q-j}v \neq 0. \end{align*} $$

This finishes the proof of the theorem.

Proof. The fact that $\mathcal {H}_{\chi }$ is a module category over $\mathscr {F}$ follows directly from Proposition 29. Since $\mathcal {H}_{\chi }$ is semisimple by definition, to show that it is a simple module category over $\mathscr {F}$ , it is enough to show that, starting from any simple object of $\mathcal {H}_{\chi }$ and tensoring it with finite-dimensional $\mathfrak {sl}_2$ -modules, we can obtain any other simple object of $\mathcal {H}_{\chi }$ as a direct summand, up to isomorphism. This claim follows by combining Proposition 29 with Theorem 31.

Proof. The isomorphism $\operatorname {\mathrm {End}}(Q(0))^{\text {op}} \cong U(\mathfrak {H})^{\mathfrak {g}}$ follows from the same argument as in the proof of Proposition 7. The inclusion $U(\mathfrak {H})^{\mathfrak {g}} \supseteq \mathbb {C}[z]$ is obvious. The converse follows easily from the following commutation relations:

(24) $$ \begin{align} [h,p^mq^n] &= (m-n) p^mq^n, \\ \nonumber [e,p^mq^n] &= n p^{m+1} q^{n-1} - \frac{n(n-1)}{2} p^{m} q^{n-2}z, \\ \nonumber [f,p^mq^n] &= m p^{m-1} q^{n+1} - \frac{m(m-1)}{2} p^{m-2} q^{n}z, \end{align} $$

which can be proved, for example, by induction.

Proof. The first claim is clear. We use it to prove the others.

For the second claim, note that $p^k$ generates a $\mathfrak {g}$ -submodule isomorphic to $L(k)$ . Since the action of $\mathfrak {g}$ preserves $Q^n :=\operatorname {span}\{p^i q^j \colon i+j \leq n\}$ , by counting dimensions, we see that $Q^n \cong \oplus _{k=0}^n L(k)$ . The claim now follows by taking colimits.

The last claim can be checked directly (enough on the generator of $Q(0)$ ).

Proof. As before, the $\mathfrak {s}$ -action on $Q(0)$ and $Q(0,\chi )$ will be denoted by $\circ $ .

Note that $[q,p^n] = -n p^{n-1}z$ and $[p,q^n] = n q^{n-1}z$ . Using this and the equations (24), one can check that

$$ \begin{align*} (pf - nq) \circ p^n = \frac{n(n+1)}{2}\chi \cdot p^{n-1}. \end{align*} $$

So, if $\chi \neq 0$ , the module $Q(0,\chi )$ is simple.

Alternatively, one can use a nilradical argument analogous to the one in Remark 13.

If $\chi =0$ , then p and q commute in $Q(0,\chi )$ , and $\mathfrak {g}$ preserves the total degree of monomials $p^iq^j$ . The rest of the proof is obvious.

Proof. This reduces to solving the equation $(\lambda _1+1)(\lambda _1+2) = (\lambda ^{\prime }_1+1)(\lambda ^{\prime }_1+2)$ .

Proof. Denote by $v_\lambda $ a basis element of $\mathbb {C}_\lambda $ . Then, $\Delta (\lambda )$ has a basis of weight vectors $\{ f^i q^j v_\lambda \colon i,j \geq 0\}$ . A direct computation shows that

$$ \begin{align*} e \cdot f^i q^j v_\lambda = i(\lambda_1 - i -j +1) f^{i-1} q^j v_\lambda + \lambda_2 \frac{j(j-1)}{2} f^i q^{j-2} v_\lambda. \end{align*} $$

This implies that the required filtration is given by the degree of q. The subquotients are given by the span of $\{ f^i q^k v_\lambda \colon i \geq 0\}$ , which is clearly isomorphic to $\Delta ^{\mathfrak {g}}(\lambda _1 - k)$ . The rest can be proved in the same way as for Lemma 16.

Proof. The left-hand side has a filtration with subquotients equal to the summands on the right-hand side. But these subquotients have different central characters by Lemma 45, since the first components of their highest weights have the same parity, so they must split.

Proof. Lemma 46, the fact that the functor $\operatorname {\mathrm {\mathbf {EA}}}$ commutes with the forgetful functor from $\mathfrak {s}$ -modules to $\mathfrak {g}$ -modules, and (the Schrödinger analogue of) Example 23 imply the decomposition (26).

Note that the lowest weight vector of a $L(\mu )$ in (26) is $f^{-1} q^t v_\lambda $ , where $t:=2+\lambda _1+\mu $ . To prove simplicity, it is enough for $\mu \geq \left | \lambda _1 + \frac {3}{2}\right | + \frac {1}{2}$ to find an element in $U(\mathfrak {s})$ that maps $f^{-1} q^t v_\lambda $ to $f^{-1} q^{t-1} v_\lambda $ . One can check by direct calculation that

$$ \begin{align*} \left( p - \frac{1}{\mu} qe \right) \cdot f^{-1} q^t v_\lambda = \frac{\lambda_2 t}{2 \mu}(\mu-\lambda_1-1) f^{-1} q^{t-1} v_\lambda. \end{align*} $$

The scalar on the right-hand side is nonzero because of the assumption on $\mu $ .

Alternatively, one can use a nilradical argument analogous to the one in Remark 25.

Proof. We use induction over n. The basis is given in Corollary 49. Suppose the proposition is true for all $k=0,\ldots ,n-1$ , where $n\geq 1$ is fixed. Observe that, using (the Schrödinger version of) Proposition 22 and Lemma 47, we have

(27) $$ \begin{align} Q(n,\lambda_2) \cong & \, Q(0,\lambda_2) \otimes L(n) \cong \operatorname{\mathrm{\mathbf{EA}}}(\Delta(-1,\lambda_2)) \otimes L(n) \\ \nonumber \cong & \operatorname{\mathrm{\mathbf{EA}}}(\Delta(n-1,\lambda_2)) \oplus \operatorname{\mathrm{\mathbf{EA}}}(\Delta(n-3,\lambda_2)) \oplus \cdots \oplus \operatorname{\mathrm{\mathbf{EA}}}(\Delta(-n-1,\lambda_2)) \\ \nonumber \cong & \, V(n,\lambda_2) \oplus V(n-2,\lambda_2) \oplus \cdots \oplus V(\epsilon,\lambda_2) \oplus{} \\ \nonumber & \oplus V(\epsilon-1,\lambda_2) \oplus V(\epsilon-3,\lambda_2) \oplus \cdots \oplus V(-n+1,\lambda_2), \end{align} $$

where $\epsilon \in \{0,1\}$ is of the same parity as n. By inductive assumption, it follows that

$$\begin{align*}\nonumber Q(n,\lambda_2) \cong & \, V(n,\lambda_2) \oplus V(n-1,\lambda_2) \oplus \cdots \oplus V(1,\lambda_2) \oplus V(0,\lambda_2) \\ \nonumber \cong & \, V(n,\lambda_2) \oplus Q(n-1,\lambda_2). \end{align*}$$

In the same way, but using $Q(0,\lambda _2) \cong \operatorname {\mathrm {\mathbf {EA}}}(\Delta (-2,\lambda _2))$ in the first line (27), we can get that $Q(n,\lambda _2) \cong V(-n,\lambda _2) \oplus Q(n-1,\lambda _2)$ . It follows that $V(n,\lambda _2) \cong V(-n,\lambda _2)$ .

Proof. The proof is analogous to the proof of Proposition 34, but easier. So we will omit it.

Proof. The fact that $\mathcal {K}_{\chi }$ is a module category over $\mathscr {F}$ follows directly from Proposition 53. Since $\mathcal {K}_{\chi }$ is semisimple by definition, to show that it is a simple module category over $\mathscr {F}$ , it is enough to show that, starting from any simple object of $\mathcal {K}_{\chi }$ and tensoring it with finite-dimensional $\mathfrak {sl}_2$ -modules, we can obtain any other simple object of $\mathcal {K}_{\chi }$ as a direct summand, up to isomorphism. This claim follows by combining Proposition 53 with Theorem 54.

Proof. Fix an element $c \in [\mathfrak {L},\mathfrak {r}] \cap \mathfrak {r}_0$ . Then, $c \in \operatorname {\mathrm {Nrad}}(\mathfrak {L})$ , so c must annihilate any simple finite-dimensional $\mathfrak {L}$ -module.

The idea is to start with the simple quotient of a Verma module for $\mathfrak {L}$ , which has finite multiplicities of its $\mathfrak {g}$ -submodules (but possibly infinite-dimensional), and then use the functors $\operatorname {\mathrm {\mathbf {C}}}_s$ to obtain an $\mathfrak {L}$ -module that should have a finite-dimensional $\mathfrak {g}$ -submodule. Then, the image of the universal $\mathfrak {L}$ -module in the constructed module should have only finite-dimensional $\mathfrak {g}$ -submodules with finite multiplicities. The element c will insure infinite-dimensionality.

Fix an antidominant, regular, and integral $\lambda \in \mathfrak {h}^\ast $ , and extend it to $\tilde {\lambda } \in \tilde {\mathfrak {h}}^\ast $ such that $\tilde {\lambda }(c) \neq 0$ . Consider the Verma module $\Delta (\tilde {\lambda })$ as in (2), and its simple quotient $\mathbf {L}(\tilde {\lambda })$ .

By considering the $\mathfrak {h}$ -weight spaces of $\Delta (\tilde {\lambda })$ (Proposition 4), it follows that as a $\mathfrak {g}$ -module, $\mathbf {L}(\tilde {\lambda })$ has a $\mathfrak {g}$ -direct summand $\Delta ^{\mathfrak {g}}(\lambda )$ generated by the highest weight vector $\mathbf {v} \in \mathbf {L}(\tilde {\lambda })$ , and this is the only $\mathfrak {g}$ -composition factor of $\mathbf {L}(\tilde {\lambda })$ of this $\mathfrak {g}$ -central character.

From the fact that $\Delta (\tilde {\lambda })$ and $\mathbf {L}(\tilde {\lambda })$ have finite-dimensional weight spaces, it follows that, when considered as $\mathfrak {g}$ -modules, they contain any simple $\mathfrak {g}$ -composition factor with at most finite multiplicity. Moreover, from this, it follows that their components in any fixed $\mathfrak {g}$ -central character lie in the category $\mathcal {O}$ for $\mathfrak {g}$ .

The element c acts on $\mathbf {v} \in \Delta ^{\mathfrak {g}}(\lambda ) \subseteq \mathbf {L}(\tilde {\lambda })$ by the scalar $\tilde {\lambda }(c) \neq 0$ .

Choose a reduced expression $w_0 = s_1 s_2 \ldots s_k$ of the longest element in the Weyl group for $\mathfrak {g}$ , and set $\operatorname {\mathrm {\mathbf {C}}}_{w_0} = \operatorname {\mathrm {\mathbf {C}}}_{s_1} \circ \operatorname {\mathrm {\mathbf {C}}}_{s_2} \circ \cdots \circ \operatorname {\mathrm {\mathbf {C}}}_{s_k}$ .

Consider for a moment $\Delta ^{\mathfrak {g}}(\lambda )$ as an $\mathfrak {L}$ -module, by declaring that $\mathfrak {r}$ acts trivially on it. From Remark 60 and [Reference Andersen and Stroppel2, Theorem 2.3], it follows that $\operatorname {\mathrm {\mathbf {C}}}_{w_0}(\Delta ^{\mathfrak {g}}(\lambda ))$ is isomorphic to the dual dominant $\mathfrak {g}$ -Verma module for the opposite Borel subalgebra. From this, we can conclude $\operatorname {\mathrm {\mathbf {C}}}_{w_0}(\Delta ^{\mathfrak {g}}(\lambda ))$ contains the finite-dimensional $\mathfrak {g}$ -submodule $L(\mu )$ , where $\mu := w_0 \cdot \lambda $ is dominant and integral. Moreover, by observing what is happening on the $\mathfrak {sl}_2$ -subalgebras of $\mathfrak {g}$ , one can conclude that the lowest weight vector in $L(\mu )$ is given by

(28) $$ \begin{align} f_1^{-1} f_2^{-1} \ldots f_k^{-1} \mathbf{v}, \end{align} $$

where $f_i$ is the negative root vector corresponding to $s_i$ .

It follows that $M:=\operatorname {\mathrm {\mathbf {C}}}_{w_0}(\mathbf {L}(\tilde {\lambda }))$ as a $\mathfrak {g}$ -module has also $L(\mu )$ as a $\mathfrak {g}$ -direct summand. Moreover, from Remark 60, it follows that $L(\mu )$ appears in M precisely once, and that each simple $\mathfrak {g}$ -module appears with at most finite multiplicity.